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I'm a mathematician who's currently teaching a course on differential equations. Though I don't know much about chemistry, I like to include examples from chemistry in my course, and I prefer for the details to be accurate. Here is a typical exam problem:

When a container of gaseous nitrogen dioxide is heated above 150°C, the gas begins to decompose into oxygen and nitric oxide: $$ 2\;\mathrm{NO}_2 \;\longrightarrow\; \mathrm{O}_2 + 2\;\mathrm{NO} $$ The rate of this reaction is determined by the equation $$ \frac{d\,[\mathrm{NO}_2]}{dt} \;=\; -k\,[\mathrm{NO}_2]^2 $$ where $k$ is a constant.

(a) Find the general solution to the above equation.

(b) A large container holds 50.0 moles of NO2 at a constant temperature of 600°C. After one hour, only 34.3 moles remain. How much NO2 will there be after another hour?

So my questions are:

  1. Is the science in this problem reasonably accurate? Is there anything you would change? (I looked up a suitable value of the rate constant $k$ to make sure that the time in part b was reasonable.)

  2. What are some other examples of reactions that are governed by simple rate laws? Ideally, I'd like to have several examples each of reactions governed by the equations $$ \frac{dy}{dt} = -ky,\qquad \frac{dy}{dt}=-ky^2,\qquad\text{and}\qquad \frac{dy}{dt}=-ky^3. $$ (Is $y^3$ really possible? Are non-integer powers of $y$ possible?)

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Check out this reference I found to answer another question: science.widener.edu/~svanbram/chem146/ch15/lm_c15_graphs.pdf It's probably well below the level that you are looking for, but it will give you an idea of how the kinetics can become more and more complicated as the number of reactants increases. –  jonsca Mar 16 at 19:34
    
This is that question, if you have a better demo or find an egregious error in mine, please leave another answer :) –  jonsca Mar 16 at 19:35
    
Since it was somehow forgotten in the joy over the nice question: Welcome to Chemistry.SE! (And yes, the idea to use real life examples is nice!) –  Klaus Warzecha Mar 17 at 8:38
    
Take a look to the wonderful iodine clock reactions I think you could find interesting this paper for the ultimate example! –  G M Mar 17 at 10:22
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2 Answers 2

up vote 10 down vote accepted

How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic.

I hope you've seen the rate equation page on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the reaction order page (you can find examples here of reactions with fraction or negative order with respect to some reagents).

Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation:

$$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$

, where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be:

$$r=k[A]^a[B]^b[C]^c[D]^d...$$

, where r is the reaction rate and k is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by:

$$\frac{d\Gamma}{dt}=\gamma r$$

However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation:

$$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$

If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant k for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong.

It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the fastest route from reactants to products, and the reaction rate is bottlenecked by the slowest step of the fastest route, which becomes responsible for determining the rate equation.

Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type $r=k[A]^3$, $r=k[A]^2[B]$, $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the focus of research when found. I don't expect anyone to know a fifth order reaction.

At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the misleading simplicity of the reaction between hydrogen and bromine.

Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as this one. The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$.

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Thank you -- this is very helpful for my overall understanding. In particular, I had been wondering why the order of a reaction is only sometimes equal to the number of reactants. –  Jim Belk Mar 16 at 21:23
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The Wikipedia pages that you mentioned are also quite helpful. The "rate equation" page lists three reactions (decompositions of hydrogen peroxide, sulfuryl chloride, and dinitrogen pentoxide) that are first order, and which I ought to be able to use. However, the only second-order reaction listed is the decomposition of nitrogen dioxide. Do you happen to know any other simple examples of second-order reactions? –  Jim Belk Mar 16 at 21:27
    
By the way, the link you gave on the hydrogen bromine reaction (with a 3/2 exponent for [$\mathrm{Br}_2$] at low $\mathrm{Br}_2$ concentrations) is quite helpful. I will certainly be using this example! –  Jim Belk Mar 16 at 21:47
    
I imagine you want a reaction that simply has at least one reagent with a second-order dependence (so $r=k[A]^2$ would work just as well as $r=k[A]^2[B]$)? I'm actually having a bit of a hard time finding a simple example of the former, even with @jlandercy's excellent source. Perhaps someone else has an example they could share? –  Nicolau Saker Neto Mar 16 at 21:56
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It's true that the former would be better, but the latter ($r=k[A]^2[B]$) would be fine as long as I could describe in words a realistic situation for which $[B]$ is much larger than $[A]$, and hence essentially constant. –  Jim Belk Mar 16 at 22:01
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I have no example in mind but fractional and high partial order (exponent) are possible. When this kind of relationship appears in chemistry they often result from complex reaction schema where more than one steps are required.

Anyway have look to Kinetic NIST Database, you will find many reaction kinetic constants and models.

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Thanks. The NIST database seems to be much more useful than my old method (Google Books) for actually looking up the reaction constants. –  Jim Belk Mar 16 at 21:32
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